Answer:
44.7 kWh
Explanation:
Let's consider the reduction of Al₂O₃ to Al in the Bayer process.
6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻
We can establish the following relations:
- The molar mass of Al is 26.98 g/mol.
- 2 moles of Al are produced when 6 moles of e⁻ circulate.
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
- 1 V = 1 J/c
- 1 kWh = 3.6 × 10⁶ J
When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:
Answer:
a) 64.27%
b) 58%
c) ethanol is the limiting reactant
d) ethanol is the limiting reactant
Explanation:
We have to note that the expected yield is the theoretical yield while the actual mass or amount of product formed is the actual yield.
a) theoretical yield=68.3g
Actual yield= 43.9 g
Percentage yield= 43.9/68.3 ×100
Percentage yield= 64.27%
b) theoretical yield= 0.0722 moles
Actual yield = 0.0419
Percentage yield= 0.0419/0.0722 × 100
Percentage yield= 58%
c) note that the limiting reactant yields the least number of moles of product
Ethanol will be the limiting reactant since it is not present in excess.
d) from the reaction equation;
1 mole of acetic acid produces 1 mole of ethyl acetate
0.58 moles of ethanol produces 0.58 moles of ethyl acetate
1 mole of acetic acid yields 1 mole of ethyl acetate
Hence 0.82 moles of acetic acid yields 0.82 moles of ethyl acetate
Hence ethanol is the limiting reactant.
Answer:
125 L of 15% and 265.625 L of 65% HCl
Explanation:
let x = volume of 15% HCl, y = 65% HCl
15x + 65y = 49 × 390.625
15x + 65y = 19140.625
x + y = 390.625
x = 390.625 - y
substitute for x in equation 1
15 (390.625 - y) + 65y = 19140.625
5859.375 - 15 y + 65 y = 19140.625
50 y = 19140.625 - 5859.375 = 13281.25
y = 132181.25 / 50 = 265.625 L
X = 390.625 - y = 390.625 - 265.625 = 125 L
https://sciencing.com/make-3d-model-atom-5887341.html